A hydrogen like atom of atomic number Z is in an excited state of quantum number 2n. It can emit a maximum energy photon of 204 eV. If it makes a transition to quantum state n, a photon of energy 40.8 eV is emitted. Find n, Z and the ground state energy (in eV) of this atom. Also calculate the minimum energy(eV) that can be emitted by this atom during de - excitation. Ground state energy of hydrogen atom is -13.6 eV

Answer

`n = 2, Z = 4, -217.6eV, 10.58 eV`

Explanation

Let ground state energy (in eV) be `E_(1)`
Then, from the given condition `E_(2n) - E_(1) = 204 eV`
`rArr (E_(1))/(4n^(2)) - E_(1) = 204 eV`
`rArr E_(1) ((1)/(4n^(2)) - 1) = 204 eV` ....(i)
and `E_(2n) - E_(n) = 40.8 eV rArr (E_(1))/(4n^(2)) - (E_(1))/(4n^(2)) = 40.8 eV`
`rArr E_(1) ((-3)/(4n^(2))) = 40.8 eV`...(ii)
From equation (i) and (ii), `(1 - (1)/(4n^(2)))/(3/(4n^(2))) = 5`
`rArr 1 = (1)/(4n^(2)) + (15)/(4n^(2)) rArr (4)/(n^(2)) = 1 rArrn = 2`
From equation number (ii),
`E_(1) = - (4)/(3) n^(2) (40.8) eV = - (4)/(3) (2)^(2) (40.8) eV`
`E_(1) = - 217.6 eV rArr E_(1) = - (13.6)Z^(2)`
`:. Z^(2) = (E_(1))/(-13.6) = (-217.6)/(-13.6) = 16 :. Z = 4`
`E_("min") = E_(2n) - E_(2n - 1)`
`= (E_(1))/(4n^(2)) - (E_(1))/((2n - 1)^(2)) = 2 = (E_(1))/(16) - (E_(1))/(9) = - (7)/(144) E_(1)`
`= - ((7)/(144)) (-217.6) eV`
`:. E_("min") = 10.58 eV`